Cut-off point of linear discriminant rule for large dimension.

Cut-off point of linear discriminant rule for large dimension

Takayuki Yamada¹, Tetsuto Himeno²and Tetsuro Sakurai³

1 General Studies, College of Engineering, Nihon University,

1 Nakagawara, Tokusada, Tamuramachi, Koriyama, Fukushima 963-8642, Japan

2Department of Computer and Information Science, Faculty of Science and Technology, Seikei University,

3-3-1 Kichijoji-Kitamachi, Musashino-shi, Tokyo 180-8633, Japan

3Center of General Education, Tokyo University of Science, Suwa,

5000-1, Toyohira, Chino-shi, Nagano, 391-0292, Japan

Abstract

This paper is concerned with the problem of classifying a observation vector into one of two pop- ulations Π1:Np(µ₁,Σ) and Π2:Np(µ₂,Σ). Anderson (1973, Ann. Statist.) gave an asymptotic expansion of the studentized statistic, and derived cut-off point to achieve a specified probability of misclassification. But the dimensionpgets large, the precision becomes worse. So in this paper, we proposed studentized statistic in terms of (n, p) asymptotic. An asymptotic expansion of the statis- tic is derived up to the orderO1, whereO1is a term with respect to{p^−1/2, N₁^−1/2, N₂^−1/2, m^−1/2} for each sample sizeNi andm=N1+N2−2−p. Using the expansion, we gave cut-off point to achieve a specified probability of misclassification.

Keywords: Linear discriminant rule, cut-off point, (n, p) asymptotic AMS 2010 subject classification: Primary 62H30, Secondary 62E20

1 Introduction

This paper is concerned with the problem of classifying a observation vector into one of two populations Π₁ :N_p(µ₁,Σ) and Π₂ :N_p(µ₂,Σ). The observation x is classified as coming from either Π₁ or Π₂ based on the samples

xi1, . . . ,xiN_i ∼Np(µ_i,Σ) (i= 1,2),

which are independent. For this problem, linear discriminant analysis is used. Let W = (¯x1−x¯2)^′S⁻¹

{ x−1

2(¯x1+ ¯x2) }

,

where ¯x1, ¯x2andS are the sample mean vectors and the pooled sample covariance matrix defined by

¯ xi= 1

Ni N_i

∑

j=1

xij, i= 1,2, S= 1 n

∑2

i=1 N_i

∑

j=1

(xij−x¯i)(xij−x¯i)^′, n=N−2 =N1+N2−2.

Linear discriminant rule classifiesxas Π₁ifW(x)> cand Π₂ifW(x)< cfor a constantc. Inference concerning linear discriminant analysis is studied under large sample asymptotic framework A0:

A0 :N₁→ ∞, N₂→ ∞, N₁/N₂→c∈(0,∞).

1E-mail address: yma801228@gmail.com

2E-mail address: t-himeno@st.seikei.ac.jp

3E-mail address: sakurai@rs.tus.ac.jp

For a review of results under A0, see, e.g., Fujikoshi et al. [4]. When p becomes large, accuracy of results proposed using A0 gets worse. As a way to improve the poorness, it is studied under the asymptotic framework A1:

A1 :p→ ∞, N1→ ∞, N2→ ∞, p/N1→γ1∈(0,1), p/N2→γ2∈(0,1) and N1/N2→γ∈(0,∞).

As results under A1, Fujikoshi and Seo [3] gave an asymptotic approximation of the probabilities of misclassification, Fujikoshi [2] gave its error bound. These results are reviewed in Fujikoshi et al. [4].

Following Lachenbruch [5], forx∈Π_i, W = (¯x1−x¯2)^′S⁻¹

{ x−1

2(¯x1+ ¯x2) }

=V^1/2Zi+ (−1)ⁱUi, (1) where

V = (¯x1−x¯2)^′S⁻¹ΣS⁻¹(¯x1−x¯2), Z_i=V^−1/2(¯x₁−x¯₂)^′S⁻¹(x−µ_i), U_i= (−1)ⁱ⁺¹(¯x₁−x¯₂)^′S⁻¹(¯x_i−µ_i)−1

2D²,

and D² is the squared sample Maharanobius distance defined by D² = (¯x1 −x¯2)^′S⁻¹(¯x1−x¯2).

From the normality of x, Zi is distributed as the standard normal distribution, which we denote it as Zi ∼ N(0,1). Since it does not depend on {x¯1,x¯2,S}, Zi is independent to the set, and so Zi

is independent from {Ui, V}. The limiting distribution of W under the asymptotic framework A1 is normal with meanui,0= (−1)ⁱlimA1E[Ui] and variancev0= limA1{E[V] + Var(Ui)}ifx∈Np(µ_i,Σ).

Under the assumption that the Mahalanobis distance ∆ =

√

(µ₁−µ2)^′Σ⁻¹(µ₁−µ2) converges to a positive constant under A1, Fujikoshi and Seo [3] and Fujikoshi [2] showed that Var(Ui)→0. So, we can abbreviate asv0= limA1E[V].

One may want to determine the cut-off point c to adjust the probabilities of misclassification.

Results under A0 is written in Anderson [1]. On the other hand, from Fujikoshi [2], under the as- sumption thatx∈Π_i, the limiting distribution (W −u_i)/√

v isN(0,1), whereu_i andv are constants such that lim_A1(u_i−u_i,0) = 0 and lim_A1(v−v₀) = 0. Using this result, we find that the approxi- mation of the misclassification probability that xis allocated to Π_j even thoughx∈ Π_i is given as Φ((−1)ⁱ⁻¹(c−u_i)/√

v) fori, j = 1,2 with i̸= j, where Φ(.) is the cumulative distribution function of the standard normal distribution. Since ui and v contain ∆², which need to be estimated. The unbiased estimator is given as

∆c²=n−p−1

n (¯x₁−x¯₂)^′S⁻¹(¯x₁−x¯₂)− pN N1N2

.

Consistency under the asymptotic framework A1 holds. One can choose c from the fact that the limiting distribution of (W−ubi)/√

ˆ

visN(0,1) ifx∈Np(µ_i,Σ), where (ubi,v) is (uˆ i, v) with replacing

∆² by∆c².

This paper is organized as follows. In Section 2, we give an asymptotic expansion of the (W−ubi)/√ ˆ v for the case thatx∈Πi. We propose a cut-off point which the misclassification probability becomes presetting level, asymptotically. Section 3 presents simulation results for misclassification probability.

Proof of lemma and derivation of expectations are given in Appendix.

2 Asymptotic expansion under A1

For technical reason, set

u_i= n 2(m−1)

{

(−1)ⁱ⁺¹∆²+ ( p

N₂ − p N₁

)}

, v= n²(n+ 1)

(m−1)(m+ 1)(m+ 2) (

∆²+ N p N1N2

) ,

where m =n−p. Note that u_i = (−1)ⁱE[U_i], but v equals E[V], asymptotically, under A1. Then lim_A1u_i=u_i,0 and lim_A1v=v₀. Using unbiased estimator of (u_i, v), we have

P (

W√−cu1

ˆ v < x

x∈Π1

)

=E [

Φ (√

ˆ

vx+√U1+cu1

V

)]

,

P (

W−cu₂

√vˆ > x x∈Π₂

)

=E [

Φ (−√

ˆ

vx+U₂−uc₂

√V

)]

. Let

u1= ( 1

N₁ + 1 N₂

)₋1/2

Σ^−1/2(¯x1−x¯2), u2= 1

√NΣ^−1/2(N1x¯1+N2x¯2−N1µ₁−N2µ₂), B=Σ^−1/2SΣ^−1/2,

where N = N₁ +N₂. Then u₁, u₂ and B are independent. In addition, u₁ ∼ N_p((1/N₁ + 1/N2)^−1/2δ,Ip) and u2 ∼ Np(0,Ip), where δ = Σ^−1/2(µ₁ −µ₂). It also holds that nB is dis- tributed as a Wishart distribution with degrees of freedomn=N−2 and covariance matrixIp, which is denoted asWp(n,Ip). Substituting them,

Ui=−(−1)ⁱ⁺¹ 2

( p N₂ − p

N₁

)u^′₁B⁻¹u1

p +(−1)ⁱ⁺¹p

√N₁N₂

u^′₁B⁻¹u2

p −τi

δ^′B⁻¹u1

√p , V = N p

N1N2

u^′₁B⁻²u1

p , τ_i=

√

pN_3/2+(−1)i+1/2

N N_3/2−(−1)i+1/2

. In addition,

∆c²= N p N1N2

{m−1 n

u^′₁B⁻¹u₁ p −1

} . Then,

b

u_i= (−1)ⁱ⁺¹ n 2(m−1)

{ N p N₁N₂

m−1 n

u^′₁B⁻¹u1

p − N p

N₁N₂ + (−1)ⁱ⁺¹ ( p

N₂ − p N₁

)}

, ˆ

v= n(n+ 1) (m+ 1)(m+ 2)

N p N1N2

u^′₁B⁻¹u1

p .

The following lemma gives that these random variables can be expressed as functions of the inde- pendent standard normal and chi-squared variables, simultaneously.

Lemma 1. Let v1∼Np(δ,Ip),v2∼Np(0,Ip),A∼Wp(n,Ip), andv1,v2 andAare independent.

Then the following equalities in distribution hold, simultaneously:

S ≡δ^′A⁻¹v1=D ∆ Y1

(

Z1+ ∆−

√Y2

Y3

Z2

) ,

T ≡v^′₂A⁻¹v₁=^D

√ 1 Y₁²

( 1 + Y₂

Y3

)

{(Z₁+ ∆)²+Z₂²+Y₄}Z₃ U ≡v^′₁A⁻¹v₁=^D 1

Y1{(Z₁+ ∆)²+Z₂²+Y₄}, V ≡v^′₁A⁻²v1=D 1

Y₁² (

1 +Y2

Y3

)

{(Z1+ ∆)²+Z₂²+Y4},

where ∆ = √

δ^′δ; Z₁, Z₂, Z₃, Y₁, . . . , Y₄ are independent, Z_i ∼ N(0,1), i = 1,2,3, Y_i ∼ χ²_f

i, i = 1, . . . ,4,

f1=n−p+ 1, f2=p−1, f3=n−p+ 2, f4=p−2.

Note that Fujikoshi and Seo [3] has also given similar results, but their results are individual ones, so cannot treat simultaneously. The proof of Lemma 1 is given in Appendix.

It can be described that (−1)ⁱ⁺¹√

b

vx+Ui−(−1)ⁱubi

= (−1)ⁱ⁺¹

√

n(n+ 1) (m+ 1)(m+ 2)ω⁻¹

√ Q1

p x−(−1)ⁱ⁺¹ 2

( p N₂ − p

N₁ )Q1

p +(−1)ⁱ⁺¹p

√N1N2

B2

p −τi

B1

√p +ω⁻²

2 Q1

p − n

2(m−1)ω⁻²+(−1)ⁱ⁺¹ 2

n m−1

( p N₂ − p

N₁ )

, (2)

V =ω⁻²Q₂

p , (3)

whereQ₁=u^′₁B⁻¹u₁,Q₂=u^′₁B⁻²u₁,B₁=δ^′B⁻¹u₁andB₂=u^′₂B⁻¹u₁,ω²=N₁N₂/(N p). From Lemma 1, we have

Q1

p

=D n f1

1 1 +√

2/f1W1

S, B1

√p

=D n f1

∆ 1 +√

2/f1W1

( Z1

√p+ω∆−

√ f2

f3

√TZ2

√p )

, B₂

p

=D n f1

1 1 +√

2/f1W1

√(

1 +f₂ f3

T )

SZ₃

√p, Q2

p

=D n² f₁²

1 (1 +√

2/f₁W₁)² (

1 +f2

f₃T )

S, whereW_i=√

f_i/2(Y_i/f_i−1) fori= 1, . . . ,4, S=

(Z1

√p+ω∆

)2

+ (Z2

√p )2

+p−2 p

( 1 +

√2 f4

W4

) , T =1 +√

2/f2W2

1 +√ 2/f3W3

.

SortingS in descending order, it can be expressed as

S=s0+S_1/2+S1+O_3/2, where

s0= 1 +ω²∆², S_1/2= 2ω∆

√p Z1+

√2 f₄W4, S1= Z₁²

p +Z₂² p −2

p,

andO_j/2is a term ofj-th order with respect to{p^−1/2, N₁^−1/2, N₂^−1/2, m^−1/2}. By Maclaurin expansion of (1 +√

2/f3W3)⁻¹ up to the term with order off₃⁻¹, T =

( 1 +

√2 f2

W2

) ( 1−

√2 f3

W3+ 2 f3

W₃² )

+O3/2,

which can be sorted in descending order as

T = 1 +T1/2+T1+O3/2

with

T_1/2=

√2 f₂W2−

√2 f₃W3, T1= 2

f3

W₃²− 2

√f2f3

W2W3. Doing Maclaurin expansion of (1 +√

2/f1W1)⁻¹ in Q1/pup to the term with order off₁⁻¹, and sort it in descending order, it is written as

Q₁

p =q1,0+Q1,1/2+Q1,1+O3/2, where

q1,0= n f₁s0, Q_1,1/2= n

f1

( S_1/2−

√2 f1

s₀W₁ )

, Q1,1= n

f1

( S1−

√2 f1

S_1/2W1+ 2 f1

s0W₁² )

, and using this expansion,

√ Q1

p =√q1,0

{ 1 + 1

2q_1,0(Q_1,1/2+Q1,1)− 1

8q_1,0² Q²_1,1/2 }

+O_3/2. Using similar way, we can expressQ2/pas

Q₂

p =q2,0+Q2,1/2+Q2,1+O3/2, where

q_2,0= (n

f1

)2( 1 +f₂

f3

) s₀, Q_2,1/2=

(n f1

)2[{(

1 + f₂ f3

)

S_1/2+f₂ f3

s₀T_1/2 }

−2

√2 f1

( 1 +f₂

f3

) s₀W₁

] , Q_2,1=

(n f₁

)2[{(

1 + f2

f₃ )

S₁+f2

f₃S_1/2T_1/2+f2

f₃s₀T₁ }

−2

√2 f₁

{(

1 + f2

f₃ )

S_1/2+f2

f₃s₀T_1/2 }

W₁ +6

f₁ (

1 + f2

f₃ )

s0W₁² ]

. In addition, it is also expanded that

B₁

√p =b1,0+B1,1/2+B1,1+O3/2, where

b_1,0= n f₁ω∆², B_1,1/2= n

f1

{(

Z₁

√p−

√ f₂ f3

Z₂

√p )

∆−

√2 f1

ω∆²W₁ }

,

B_1,1= n f₁

{

−∆ 2

√ f2

f₃ Z2

√pT_1/2−

√2 f₁

( Z1

√p−

√ f2

f₃ Z2

√p )

∆W₁+ 2

f₁ω∆²W₁² }

.

Sorting in descending order, it can be described that (

1 + f2

f₃T )

S=s0

( 1 +f2

f₃ )

(1 + ˜T_1/2+ ˜S_1/2) +O1, where ˜S_1/2=S_1/2/s0 and ˜T_1/2={(f2/f3)/(1 +f2/f3)}T_1/2, and so

√(

1 +f₂ f3

T )

S=

√ s0

( 1 + f₂

f3

) ( 1 + 1

2( ˜T1/2+ ˜S1/2) )

+O1. Substituting this expansion intoB2/p, and using Maclaurin expansion of (1 +√

2/f1W1)⁻¹ up to the term with order off₁^−1/2, we have

B2

p =B_2,1/2+B2,1+O_3/2, where

B2,1/2= n f1

√(

1 + f₂ f3

) s0

Z₃

√p,

B2,1= n f₁

√(

1 +f2

f₃ )

s0

{1

2( ˜T_1/2+ ˜S_1/2)−

√2 f₁W1

} Z3

√p.

Substituting these expansions in (2) and (3), and coordinating it in order, (2) and (3) can be represented as the sum of terms with descending order, respectively, which are

(−1)ⁱ⁺¹√ ˆ

vx+Ui−(−1)ⁱubi= (−1)ⁱ⁺¹

√

n²(n+ 1)

(m+ 1)²(m+ 2)s0ω⁻¹x+Ui,1/2+Ui,1+O3/2, V =ω⁻² n²(n+ 1)

(m+ 1)²(m+ 2)s0+ω⁻²Q_2,1/2+ω⁻²Q2,1+O_3/2, where

U_i,1/2=AiQ_1,1/2−τiB_1,1/2+(−1)ⁱ⁺¹p

√N1N2

B_2,1/2, Ui,1=AiQ1,1−(−1)ⁱ⁺¹

√

n(n+ 1) (m+ 1)(m+ 2)

ω⁻¹x 8q^3/2_1,0

Q²_1,1/2+(−1)ⁱ⁺¹p

√N1N2

B2,1−τiB1,1

+ n

(m−1)(m+ 1) [

(−1)ⁱ⁺¹ ( p

N2 − p N1

)

−ω⁻² ]

, Ai= (−1)ⁱ⁺¹

√

n(n+ 1) (m+ 1)(m+ 2)

ω⁻¹

2√q_1,0x−(−1)ⁱ⁺¹ 2

( p N₂ − p

N₁ )

+ω⁻² 2 . These expansions lead that

Ri= (−1)ⁱ⁺¹√ ˆ

vx+√Ui+ (−1)ⁱ⁺¹ubi

V = (−1)ⁱ⁺¹x+Ri,1/2+Ri,1+O3/2,

where

R_i,1/2=−1

2(−1)ⁱ⁺¹xQ˜_2,1/2+ ˜U_i,1/2, Ri,1= (−1)ⁱ⁺¹x

(3 8

Q˜²_2,1/2−1 2

Q˜2,1

)

−1 2

U˜i,1/2Q˜2,1/2+ ˜Ui,1

with

U˜_i,j/2=U_i,j/2 / {√

n²(n+ 1)

(m+ 1)²(m+ 2)s₀ω⁻¹ }

, Q˜_i,j/2=Q_i,j/2

/ { n²(n+ 1) (m+ 1)²(m+ 2)s0

} . By Taylor expansion,

Φ(Ri) = Φ((−1)ⁱ⁺¹x) +ϕ((−1)ⁱ⁺¹x)[

R_i,1/2+Ri,1

]−(−1)ⁱ⁺¹x

2 ϕ((−1)ⁱ⁺¹x)R²_i,1/2+O_3/2. SinceR_i,1/2 is represented as the linear combination of{Z1, Z2, Z3, W1, . . . , W4}, E[R_i,1/2] = 0. As a result, we give the following theorem.

Theorem 1. Let

˜

x_i=x−{

(−1)ⁱ⁺¹E[R\_i,1]−x

2E[R\²_i,1/2] }

, whereE[R\^k_i,j] isE[R^k_i,j] with replacing∆² by ∆c². Fori= 1,2,

P (

(−1)ⁱ⁺¹W−ub_i

√vˆ <(−1)ⁱ⁺¹x˜i

x∈Πi

)

= Φ((−1)ⁱ⁺¹x) +O3/2. Explicit formula ofE[R_i,1] andE[R²_i,1/2] can be derived, which are given in Appendix.

Based on the expansion, we set cutoff pointci as c_i=√

ˆ v [

(−1)ⁱ⁺¹z_α− {

(−1)ⁱ⁺¹E[R\_i,1]−(−1)ⁱ⁺¹zα

2 E[R\²_i,1/2] }]

+ ˆu_i (i= 1,2),

wherez_αis theαpercentile point of the standard normal distribution. The cutoff pointc₁(c₂) makes the desired misclassification probability to be α within the error O_3/2. The other misclassification probabilities can be described as

P(W > c1|x∈Π2) =E [

Φ

(−(c₁−cu₂) +U₂−uc₂

√V

)]

, P(W < c₂|x∈Π₁) =E

[ Φ

((c₂−cu₁) +U₁+cu₁

√V

)]

, respectively. Note that

ˆ

v/V →^p 1, U1+ ˆu1

→p 0, U2−uˆ2

→p 0, E[R_i,j/2] =O_j/2 (i, j= 1,2).

From the expansion, we have c

u1−cu2= n m−1

{ N p N1N2

m−1 n

u^′₁B⁻¹u1

p − N p N1N2

}

=ω⁻²(1 +ω²∆²) n

m+ 1 − n

m−1ω⁻²+O_1/2. It will be found that

c

u₁−cu₂− n

m∆²→^p 0 under the asymptotic framework A1. In addition,

ˆ v− n³

m³ω⁻²(1 +ω²∆²)→^p 0.

Combining these results,

limA1 P(W > c1|x∈Π2) = lim

A1P(W < c2|x∈Π1) = Φ (

z1−α−lim

A1

√m n

∆²

√∆²+ω⁻² )

.

A Proof of Lemma 1

Proof of Lemma1. LetΓbe a orthogonal matrix of orderpwhich the first row is proportional toδ^′, and letB=ΓAΓ^′ andwi =Γvi,i= 1,2. ThenB∼Wp(n,Ip),w1∼Np(∆e1,Ip),w2∼Np(0,Ip) andw1,w2 andB are independent;

S= (Γδ)^′(ΓAΓ^′)⁻¹(Γv1)= ∆e^{D ′}₁B⁻¹w1, T= (Γv₂)^′(ΓAΓ^′)⁻¹(Γv₁)=^Dw^′₂B⁻¹w₁=^D

√

w^′₁B⁻²w₁Z U = (Γv1)^′(ΓAΓ^′)⁻¹(Γv1)=^Dw^′₁B⁻¹w1,

V = (Γv₁)^′(ΓAΓ^′)⁻²(Γv₁)=^D w^′₁B⁻²w₁,

where ei denotes fundamental vector with 1 in i-th position, Z ∼ N(0,1), and Z and {B,w1} are independent. By using reflection matrix(Householder matrix)Hbetweene1 and (1/√

w^′₁w1)w1, S = ∆^D √

w^′₁w1(He1)^′(HBH^′)⁻¹{H(1/√

w^′₁w1)w1}= ∆w^′₁(HBH^′)⁻¹e1. Besides,

U =^D w^′₁B⁻¹w1=w^′₁w1·e^′₁(HBH^′)⁻¹e1, V =^D w^′₁B⁻²w₁=w^′₁w₁·e^′₁(HBH^′)⁻²e₁. Givenw1, C≡HBH^′ ∼Wp(n,Ip), soC andw1 are independent. Partition

C=

(c11 c^′₂₁ c21 C22

)

and w1= (w11

w21

) . It can be expressed that

S= ∆w^{D ′}₁C⁻¹e1= ∆ c11·2

(w11−w^′₂₁C⁻₂₂¹c21), wherec_11·2=c₁₁−c^′₂₁C⁻₂₂¹c₂₁. In addition,

U =^D w^′₁w₁·e^′₁C⁻¹e₁= w^′₁w₁ c11·2

, V =^D w^′₁w₁·e^′₁C⁻²e₁= w^′₁w₁

c²_11·2 (1 +c^′₂₁C⁻₂₂¹c₂₁).

It is noted thatx≡C⁻₂₂^1/2c21∼Np−1(0,Ip),D≡C22∼Wp−1(n,Ip−1), andxandD are indepen- dent, thusw11,w21,x,D andc11·2 are independent. Using these results, we have

S =^D ∆ c11·2

(w₁₁−w^′₂₁D^−1/2x) and V =^D w^′₁₁w₁₁

c²_11·2 (1 +x^′D⁻¹x).

LetG be orthogonal matrix of order p−1 which the first row is proportional to x^′D^−1/2. Givenx and D, y ≡ Gw21 ∼ Np−1(0,Ip−1), and it is found that w11, c11·2, x, D and y are independent.

Partitioningy= (y1y^′₂)^′, we have S=^D ∆

c11·2{w11−(Gw21)^′(GD^−1/2x)}=^D ∆ c11·2

(w11−√

x^′D⁻¹xy1), U =^D w₁₁² + (Gw₂₁)^′(Gw₂₁)

c11·2

=D 1 c11·2

(w²₁₁+y₁²+y^′₂y₂), V =^D w²₁₁+ (Gw₂₁)^′(Gw₂₁)

c²_11·2 (1 +x^′D⁻¹x)=^D 1

c²_11·2(1 +x^′D⁻¹x)(w²₁₁+y²₁+y^′₂y₂).

These show the conclusion of the lemma.

B Expectations

Firstly, we calculateE[R²_i,1/2]. It is that E[R²_i,1/2] = x²

4 E[ ˜Q²_2,1/2] +E[ ˜U_i,1/2² ]−(−1)ⁱ⁺¹xE[ ˜Q_2,1/2·U˜_i,1/2].

SinceS_1/2⊥⊥T_1/2⊥⊥W₁, E[Q²_2,1/2] = n⁴

f₁⁴ [(

1 + f2

f₃ )2

E[S_1/2² ] + (f2

f₃ )2

s²₀E[T_1/2² ] + 4· 2 f₁

( 1 + f2

f₃ )2

s²₀E[W₁²] ]

. Noting that

S_1/2² = 4ω²∆²

p Z₁²+ 2 f4

W₄²+ 22ω∆

√p

√2 f4

Z₁W₄, it holds that

E[S_1/2² ] = 4ω²∆² p + 2

f₄. It also holds that

E[T_1/2² ] = 2 f2

E[W₂²] + 2 f3

E[W₃²] = 2 f2

+ 2 f3

. Thus,

E[Q²_2,1/2] = n⁴ f₁⁴

[(

1 +f2

f₃ )2(

4ω²∆² p + 2

f₄ )

+ (f2

f₃ )2

(1 +ω²∆²)² (2

f₂ + 2 f₃

)

+8 f1

( 1 + f2

f3

)2

(1 +ω²∆²)² ]

. For evaluatingE[U_i,1/2² ], note that

Q²_1,1/2=n² f₁²

(

S_1/2² + 2

f₁s²₀W₁²−2

√2

f₁s0S_1/2W1

) . Thus,

E[Q²_1,1/2] = n² f₁²

[(4ω²∆² p + 2

f4

) + 2

f1

s²₀ ]

= n² f₁²

[2 f4

+ 2 f1

+ (4

p+ 2 f1

) ω²∆²

] . Moreover, the following equalities hold.

E[B²_2,1/2] = n² f₁²s0

( 1 +f2

f₃ )1

p =n² f₁²

( 1 + f2

f₃ )

(1 +ω²∆²)1 p, E[B²_1,1/2] = n²

f₁² {(1

p+f₂ f3

1 p

)

ω∆²+ 2 f1

ω²∆⁴ }

. From independence,

E[B_1,1/2·B_2,1/2] =E[B_1,1/2]·E[B_2,1/2] = 0, E[Q_1,1/2·B_2,1/2] =E[Q_1,1/2]·E[B_2,1/2] = 0.